# Chemistry

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Given a saturated solution of CaF2, write the ionic equilibrium and use it to calculate the solubility of CaF2.

My answer for the ionic equation is
CaF2(s)<->Ca2+(aq)+ 2F-(aq)
Ksp= [Ca2+][F-]2

the given ksp is 3.32*10-4

how do i figure out the solubility??
i know the stoichiometry is 1:2 but i don't know how to finish this problem... please help

• Chemistry - ,

Everything you've done so far is very good.
..........CaF2 ==> Ca^2+ + 2F^-
I........solid......0.......0
C........solid.......x......2x
E........solid......x.......2x

Ksp = [Ca^2+][F^-]^2 = 3.32E-4
3.32E-4 = (x)(2x)^2 and solve for x.
x = cncn Ca^2+ in mols/L. The question asks for solubility of CaF2 but since 1 mol Ca^2+ is obtain for each CaF2 solid molecule that dissolves, then x = solubility CaF2 (in M or mols/L)

• Chemistry - ,

ksp=(x)(2x)2=3.32*10^-4
= (x)(4x2)=3.32*10^-4
= 4x3=3.32*10^-4
then i took the cubed root of that and got 2.02*10^-4???

is that correct?

• Chemistry - ,

for this same question it also asks...

what will happen to the value of ksp?
and
Calculate the new solubility of CaF2 (in 0.100mol/L NaF)

how do i go about doing the second part?

the ksp will increase beacuse you are adding more to the solution?

• Chemistry - ,

I agree down to 4x^3 = 3.32E-4. However, when I divide by 4 and take the cube root of the result I don't get your value.
When 0.1M NaF is added the solubility decreases because you now have a common ion of F^-
Here 3.32E-4 = (x)(2x+0.1)
and solve for x.

NO. NO. NO. Ksp does NOT change, K is a constant. Why would we call it a constant if it were not constant. Why does it not change? Adding more SOLID CaF2 will not make more of it dissolve. The amount that will dissolve will dissolve regardless of whether you add another spoonful of solid or not. By the way, raising or lowering the T WILL change K because the solubility changes with changing temperature.